Prove that for any set $A$, $A = \bigcup \mathscr P (A)$. Not a duplicate of
Prove that $ (\forall A)\bigcup\mathcal P(A) = A$
Prove that for any set A, A = $\cup$ $\mathscr{P}$(A)
This is exercise $3.4.16$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Prove that for any set $A$, $A = \bigcup \mathscr P (A)$.
Here is my proof:
Suppose $A$ is arbitrary.
$(\rightarrow)$ Let $x$ be an arbitrary element of $A$. Since $A\subseteq A$ then $A\in\mathscr P(A)$. From $A\in\mathscr P(A)$ and $x\in A$, $x\in\bigcup \mathscr P(A)$. Therefore if $x\in A$ then $x\in\bigcup \mathscr P(A)$. Since $x$ is arbitrary, $\forall x\Bigr(x\in A\rightarrow x\in\bigcup\mathscr P(A)\Bigr)$ and so $A\subseteq \bigcup\mathscr P(A)$.
$(\leftarrow)$ Let $x$ be an arbitrary element of $\bigcup\mathscr P(A)$. So we can choose some $A_0$ such that $A_0\in\mathscr P(A)$ and $x\in A_0$. $A_0\in\mathscr P(A)$ is equivalent to $A_0\subseteq A$ and since $x\in A_0$, $x\in A$. Therefore if $x\in\bigcup \mathscr P(A)$ then $x\in A$. Since $x$ is arbitrary, $\forall x\Bigr(x\in\bigcup\mathscr P(A)\rightarrow x\in A\Bigr)$ and so $\bigcup\mathscr P(A)\subseteq A$.
From $A\subseteq \bigcup\mathscr P(A)$ and $\bigcup\mathscr P(A)\subseteq A$ we obtain $A= \bigcup\mathscr P(A)$. Since $A$ is arbitrary, $\forall A\Bigr(A=\bigcup\mathscr P(A)\Bigr)$. $Q.E.D.$
Is my proof valid$?$
Thanks for your attention.
 A: The proof is valid but you could make it shorter and clearer by arguing at the level of sets, rather than elements.  For example $A\in\mathscr{P}(A)$ implies $A\subseteq\bigcup\mathscr{P}(A)$. And "$X\subseteq A$ for all $X\in \mathscr{P}(A)$" implies $\bigcup\mathscr{P}(A)\subseteq A$.
A: All you're really trying to do is show that given a set $A$, $A$ is the union of all of its subsets. Well, since $A$ is a subset of $A$, the result automatically follows.
A: By definition, $x \in \bigcup F$ if and only if there is some $Y \in F$ such that $x \in Y \in F$. In particular, $x \in \bigcup \mathcal{P}(A)$ if and only if there is some $Y \in \mathcal{P}(A)$, i.e. $Y \subseteq A$, such that $x \in Y$. But of course if $x \in Y \subseteq A$, then $x \in A$, and conversely, if $x \in A$, then $x \in \{ x \} \subseteq A$. So $A = \bigcup \mathcal{P}(A)$.
A: Assume the union of the powerset of A has an element not in A; then one of A's subsets has an element not in A, a contradiction. Therefore the union of the powerset of A is a subset of A.
Now assume A has an element not in the union of the powerset of A. Well in particular the element is found in A so it is in some subset.
